Question

Derive an expression for the fraction of dissociation of a diatomic gas molecule into two gaseous atoms?

Answer 1

Suppose we had a simple dissociation reaction:

`AB(g) -> A(g) + B(g)`

Then the expression for `K_P` would be:

`K_P = (P_A P_B)/(P_(AB))`

where `P_i` is the partial pressure of `i` at equilibrium.

`AB` would dissociate so that its partial pressure decreases by `x` and the partial pressures of `A` and `B` would be `x` each...

`=> K_P = x^2/(P_(AB) - x)`

The fraction of dissociation is defined by

`alpha = x/P_(AB)`.

Thus...

`K_P = (alphaP_(AB))^2/(P_(AB) - alphaP_(AB))`

`= (alpha^2P_(AB)^2)/((1 - alpha)P_(AB))`

`= (alpha^2)/(1 - alpha)P_(AB)`

And so...

`K_P - K_Palpha = P_(AB)alpha^2`

`P_(AB)alpha^2 + K_Palpha - K_P = 0`

This becomes a quadratic equation in `alpha`:

`color(blue)(alpha) = (-K_P pm sqrt(K_P^2 - 4(P_(AB))(-K_P)))/(2P_(AB))`

`= (-K_P pm sqrt(K_P^2 + 4K_PP_(AB)))/(2P_(AB))`

`= -K_P/(2P_(AB)) pm (sqrt(K_P^2 + 4K_PP_(AB)))/(2P_(AB))`

`= -K_P/(2P_(AB)) pm sqrt(K_P^2/(4P_(AB)^2) + K_P/P_(AB))`

`= -K_P/(2P_(AB)) pm K_P/(P_(AB))sqrt(1/4 + P_(AB)/K_P)`

`= color(blue)(K_P/(P_(AB))[-1/2 pm sqrt(1/4 + P_(AB)/K_P)])`

I think that's about as simple as I can get it...